Let $S = a+b$ and $P = ab$. Then $(a+1)(b+1) = ab + a + b + 1 = P + S + 1$ and $a^3 + b^3 = (a+b)(a^2 - ab + b^2) = S(S^2 - 3P)$. So
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$\left| \right.$
$$\begin{array}{l} S(P + S + 1) = 2 \\ S(S^2 - 3P) = 1 \end{array}$$
$\left. \right. \iff$
$$\begin{array}{l} 3SP + 3S^2 + 3S + S^3 - 3PS = 3 \cdot 2 + 1 \\ S^3 + 3S^2 + 3S = 7 \\ (S + 1)^3 = 8 \iff S = 1. \end{array}$$